a) Ta có: \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}=\frac{4.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}{5.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}=\frac{4}{5}\)

b) \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-.....-\frac{331}{1000.1331}\)

\(=\frac{1330}{1331}-\left(\frac{8-7}{1.8}+\frac{27-8}{8.27}+.....+\frac{1331-1000}{1000.1331}\right)\)

\(=\frac{1330}{1331}-\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{27}+....+\frac{1}{1000}-\frac{1}{1331}\right)\)

\(=\frac{1330}{1331}-\left(1-\frac{1}{1331}\right)\)

\(=\frac{1330}{1331}-\frac{1330}{1331}=0\)

Vậy \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-....\frac{331}{1000.1331}=0\)

CHÚC BẠN HỌC TỐT

a) \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}\)

\(=\frac{4\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}{5\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}\)

\(=\frac{4}{5}\)

 bÀI LÀM

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

                                      \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{13}{9}:13:\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\left(\frac{21}{315}+\frac{9}{315}+\frac{5}{315}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\frac{35}{315}=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\frac{1}{9}=-10\)

                                 <=> \(\frac{3}{5}x-1=-10\)

                                 <=> \(\frac{3}{5}x=-9\)

                                 <=> \(x=-15\)

Vậy x = -15.

\(\frac{3}{5}x-1\frac{4}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}=-10\right)\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{3.5}+\frac{2}{.57}+\frac{2}{7.9}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{1}{3}-\frac{1}{9}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{2}.\frac{2}{9}\right)=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\frac{26}{18}=-10\)

\(\Leftrightarrow\frac{3}{5}x-1=-10\)

\(\Leftrightarrow\frac{3}{5}x=-10+1\)

\(\Leftrightarrow\frac{3}{5}x=-9\)

\(\Rightarrow x=-9:\frac{3}{5}\)

\(\Rightarrow x=-15\)

Vậy \(x=-15\)

1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200

 

A=1/2+1/3+1/6+1/12
A=(1/2+1/3+1/6)+1/12
A=1+1/12
A=13/12

B=20/3+61/8+10/3+37/8
B=30/3+98/8
B=10+49/4
B=89/4

\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}=-5\right)\)

\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\right)=-5\)

\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]=-5\)

\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]=-5\)

\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]=-5\)

\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{11}\right)\right]=-5\)

\(\frac{3}{2}.x-\frac{45}{2}=-5\)

\(\frac{3}{2}.x=\frac{35}{2}\)

\(x=\frac{35}{3}\)

bài phát sai òi hihi cần mk chỉ chỗ sai ko

x= 80/3 đúng ko các bạn

Rút gọn biểu thức S, ta có:
\(S=\frac{13}{30}+\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(\Leftrightarrow S=\frac{13}{30}+\left(\frac{13}{3\cdot5}+\frac{13}{5\cdot7}+\frac{13}{7\cdot9}+\frac{13}{9\cdot11}\right)\)
Đặt \(P=\frac{13}{3\cdot5}+\frac{13}{5\cdot7}+\frac{13}{7\cdot9}+\frac{13}{9\cdot11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3}-\frac{2}{11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{16}{33}\)
\(\Rightarrow P=\frac{104}{33}=3\frac{5}{33}\)
Ta có: \(P+1>P+\frac{13}{30}\)
Mà  \(P+\frac{13}{30}=S\)
Còn \(P+1=3\frac{5}{33}+1=4\frac{5}{33}<5\)
\(\Rightarrow S<4\frac{5}{33}<5\)

Vậy đề bài sai.

Bài này sai đề rồi.

Sai rõ 

\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)

\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)

Cho tam giác ABC có đường cao AD .Gọi E là trung điểm  của AB .F đối xứng vs D qua E c/m AB = DF